Download presentation

Presentation is loading. Please wait.

Published byMisael Anthony Modified over 3 years ago

1
1 Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo.com

2
2 Portal frames

3
3 Learning out Come Review of Design of Portal Frames Design example Continued

4
4 INTRODUCTION Step1: Design of slabs Step2: Preliminary design of beams and columns Step3: Analysis Step4: Design of beams Step5: Design of Columns Step6: Design of footings

5
5 PROBLEM 2

6
6 A portal frame hinged at base has following data: Spacing of portal frames = 4m Height of columns = 4m Distance between column centers = 10m Live load on roof = 1.5 kN/m2 RCC slab continuous over portal frames. Safe bearing capacity of soil=200 kN/m2 Adopt M-20 grade concrete and Fe-415 steel. Design the slab, portal frame and foundations and sketch the details of reinforcements.

7
7 Data given: Spacing of frames = 4m Span of portal frame = 10m Height of columns = 4m Live load on roof = 1.5 kN/m 2 Concrete: M20 grade Steel: Fe 415

8
8

9
9

10
10 Step1:Design of slab Assume over all depth of slab as 120mm and effective depth as 100mm Self weight of slab = 0.12 x 24 = 2.88 kN/m2 Weight of roof finish = 0.50 kN/m2 (assumed) Ceiling finish= 0.25 kN/m2 (assumed) Total dead load wd= 3.63 kN/m2 Live load wL= 1.50 kN/m2 (Given in the data) Maximum service load moment at interior support = = 8.5 kN-m

11
11 Step1:Design of slab (Contd) Mu lim =Q lim bd2 (Q lim =2.76) = 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > 12.75 kN-m From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x 100)/100= 384 mm2 Spacing of 10 mm dia bars = (78.54 x 1000)/384= 204.5 mm c/c Provide #10 @ 200 c/c

12
12 Step1:Design of slab (Contd) Area of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm 2 Spacing of 8 mm dia bars = (50.26 x 1000)/144= 349 mm c/c Provide #8 @ 340 c/c. Main and dist. reinforcement in the slab is shown in Fig

13
13 Step1:Design of slab (Contd)

14
14 Step2: Preliminary design of beams and columns Beam: Effective span = 10m Effective depth based on deflection criteria = 10000/13 = 769.23mm Assume over all depth as 750 mm with effective depth = 700mm, breadth b = 450mm and column section equal to 450 mm x 600 mm.

15
15 Step3: Analysis Load on frame i) Load from slab = (3.63+1.5) x 4 =20.52 kN/m ii) Self weight of rib of beam = 0.45x0.63x24= 6.80 kN/m Total 28.00 kN/m Height of beam above hinge = 4+0.1-075/2 =3.72 m The portal frame subjected to the udl considered for analysis is shown in Fig. 6.10

16
16 Step3: Analysis (Contd.)

17
17 Step3:Analysis (Contd) The moments in the portal frame hinged at the base and loaded as shown in Fig. is analised by moment distribution IAB = 450 x 600 3 /12 = 81 x 10 8 mm4, IBC= 450 x 750 3 /12 = 158.2 x 10 8 mm4 Stiffness Factor: KBA= IAB / LAB = 21.77 x 10 5 KBC= IBC / LBC = 15.8 x 10 5

18
18 Step3:Analysis (Contd) Distribution Factors: Fixed End Moments: MFAB= MFBA= MFCD= MFDC 0 MFBC= -=-233 kN-m and MFCB= =233 kN-m

19
19 Step3:Analysis (Contd) Moment Distribution Table

20
20 Step3:Analysis (Contd) Bending Moment diagram

21
21 Step3:Analysis (Contd) Design moments: Service load end moments: M B =156 kN-m, Design end moments M uB =1.5 x 156 = 234 kN-m, Service load mid span moment in beam= 28x10 2 /8 – 102 =194 kN-m Design mid span moment Mu+=1.5 x 194 = 291 kN-m Maximum Working shear force (at B or C) in beam = 0.5 x 28 x 10 = 140kN Design shear force Vu = 1.5 x 140 = 210 kN

22
22 Step4:Design of beams: The beam of an intermediate portal frame is designed. The mid span section of this beam is designed as a T-beam and the beam section at the ends are designed as rectangular section. Design of T-section for Mid Span : Design moment Mu=291 kN-m Flange width bf= Here Lo=0.7 x L = 0.7 x 10 =7m bf= 7/6+0.45+6x0.12=2.33m

23
23 Step4:Design of T-beam: b f /b w =5.2 and D f /d =0.17 Referring to table 58 of SP16, the moment resistance factor is given by K T =0.43, M ulim =K T bwd 2 fck = 0.43 x 450 x 700 2 x 20/1x10 6 = 1896.3 kN-m > Mu Safe The reinforcement is computed using table 2 of SP16

24
24 Step4:Design of T- beam: Mu/bd 2 = 291 x 10 6 /(450x700 2 ) 1.3 for this p t =0.392 A st =0.392 x 450x700/100 = 1234.8 mm 2 No of 20 mm dia bar = 1234.8/( x20 2 /4) =3.93 Hence 4 Nos. of #20 at bottom in the mid span

25
25 Step4:Design of Rectangular beam: Design moment M uB =234 kN-m M uB /bd 2 = 234x106/450x700 2 1.1 From table 2 of SP16 p t =0.327 A st =0.327 x 450 x 700 / 100 = 1030 No of 20 mm dia bar = 1030/( x20 2 /4) =3.2 Hence 4 Nos. of #20 at the top near the ends for a distance of o.25 L = 2.5m from face of the column as shown in Fig

26
26 Step4:Design of beams Long. Section:

27
27 Step4:Design of beams Cross-Section:

28
28 Check for Shear: Nominal shear stress = pt=100x 1256/(450x700)=0.39 0.4 Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcement is required to be designed Strength of concrete Vuc=0.432 x 450 x 700/1000 = 136 kN Shear to be carried by steel Vus=210-136 = 74 kN

29
29 Check for Shear: Nominal shear stress = p t =100x 942/(400x600)=0.39 0.4 Permissible stress for pt=0.4 from table 19 c =0.432 < v Hence shear reinforcement is required to be designed Strength of concrete V uc =0.432 x 400 x 600/1000 = 103 kN Shear to be carried by steel V us =162-103 = 59 kN

30
30 Check for Shear: Spacing 2 legged 8 mm dia stirrup sv= Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

31
31 Step5:Design of Columns: Cross-section of column = 450 mm x 600 mm Ultimate axial load Pu=1.5 x 140 = 210 kN (Axial load = shear force in beam) Ultimate moment Mu= 1.5 x 156 = 234 kN-m ( Maximum) Assuming effective cover d’ = 50 mm; d’/D 0.1

32
32

33
33 Step5:Design of Columns: Referring to chart 32 of SP16, p/fck=0.04; p=20 x 0.04 = 0.8 % Equal to Minimum percentage stipulated by IS456-2000 (0.8 % ) Ast=0.8x450x600/100 = 2160 mm2 No. of bars required = 2160/314 = 6.8 Provide 8 bars of #20

34
34 Step5:Design of Columns: 8mm diameter tie shall have pitch least of the following Least lateral dimension = 450 mm 16 times diameter of main bar = 320 mm 48 times diameter of tie bar = 384 300mm Provide 8 mm tie @ 300 mm c/c

35
35 600 Tie #8 @300 c/c 8-#20 450

36
36 Step6:Design of Hinges: At the hinge portion, concrete is under triaxial stress and can withstand higher permissible stress. Permissible compressive stress in concrete at hinge= 2x0.4f ck =16 MPa Factored thrust =P u =210kN Cross sectional area of hinge required = 210x10 3 /16=13125 mm 2 Provide concrete area of 200 x100 (Area =20000mm 2 ) for the hinge

37
37 Step6:Design of Hinges: Shear force at hinge = Total moment in column/height = 156/3.72=42 Ultimate shear force = 1.5x42=63 kN Inclination of bar with vertical = = tan -1 (30/50) =31 o Ultimate shear force = 0.87 f y A st sin Provide 4-#16 (Area=804 mm 2 )

38
38 Step7:Design of Footings: Load: Axial Working load on column = 140 kN Self weight of column =0.45 x 0.6 x3.72x 24 = 24 Self weight of footing @10% = 16 kN Total load = 180 kN Working moment at base = 42 x 1 =42 kN-m

39
39 Step6:Design of Footings: Approximate area footing required = Load on column/SBC = 180/200 =0.9 m 2 However the area provided shall be more than required to take care of effect of moment. The footing size shall be assumed to be 1mx2m (Area=2 m 2 )

40
40 2mX 1m 0.7m 0.6m X 0.45m

41
41 Step6:Design of Footings: Maximum pressure qmax=P/A+M/Z = 180/2+6x42/1x22 = 153 kN/m 2 Minimum pressure qmin =P/A-M/Z = 180/2-6x42/1x2 2 = 27 kN/m 2 Average pressure q = (153+27)/2 = 90 kN/m 2 Bending moment at X-X = 90 x 1 x 0.72/2 = 22 kN-m Factored moment Mu 33 kN-m

42
42 Step6:Design of Footings: Over all depth shall be assumed as 300 mm and effective depth as 250 mm, Corresponding percentage of steel from Table 2 of SP16 is pt= 0.15% > Minimum p t =0.12%

43
43 Step6:Design of Footings: Area of steel per meter width of footing is A st =0.12x1000x250/100=300 mm2 Spacing of 12 mm diameter bar = 113x1000/300 = 376 mm c/c Provide #12 @ 300 c/c both ways

44
44 Step6:Design of Footings: Length of punching influence plane = a o = 600+250 = 850 mm Width of punching influence plane = b o = 450+250 = 700 mm Punching shear Force = V punch =180-90x(0.85x0.7)=126.5 kN Punching shear stress punch =V punch /(2x(a o +b o )d) =126.5x103/(2x(850+700)250) = 0.16 MPa Permissible shear stress = 0.25 fck=1.18 MPa > punch Safe

45
45 Step6:Design of Footings: Check for One Way Shear Shear force at a distance ‘d’ from face of column V= 90x1x0.45 = 40.5 kN Shear stress v =40.5x103/(1000x250)=0.162 MPa For p t =0.15, the permissible stress c = 0.28 (From table 19 of IS456-2000) Details of reinforcement provided in footing is shown in Fig.

46
46

47
47

48
48 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo.com

Similar presentations

Presentation is loading. Please wait....

OK

Dr Badorul Hisham Abu Bakar

Dr Badorul Hisham Abu Bakar

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google